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Frequency response INA

  1. Nov 8, 2017 #1
    1. QUESTION

    i'm dealing with a 3op instrumentation amplifier.

    say i have a circuit gain of 100

    and i have an INA with a gain bandwidth product of 2MHz.

    the two buffer differential amplifiers are at 100k

    the open loop gain is then 2MHz/100k = 20

    this frequency of 100k is going to effect the gain of the circuit because the open loop gain is much smaller than the gain of the circuit at 100.

    my question is: why does the open loop gain have to be larger than the circuit gain for it to not affect it?


     
    Last edited by a moderator: Nov 8, 2017
  2. jcsd
  3. Nov 8, 2017 #2

    berkeman

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    Staff: Mentor

    Can you post a Bode Plot of the open loop gain, and then superimpose the closed-loop gain on the plot? Does that help to answer your question? :smile:
     
  4. Nov 8, 2017 #3

    berkeman

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    Do you mean closed loop gain?
     
  5. Nov 8, 2017 #4
    no, i mean open loop gain. i'm looking at a graph with Acl and Aol but i still don't understand my question
     
  6. Nov 8, 2017 #5

    berkeman

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    Oh, I get what you are saying now, you mean Aol at 100kHz. Got it.
    Can you Upload it?
     
  7. Nov 8, 2017 #6

    berkeman

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  8. Nov 8, 2017 #7
    yea exactly like that
     
  9. Nov 8, 2017 #8

    berkeman

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    The open look gain has to be larger or equal to the closed loop gain that you are setting, or the closed loop gain will be less than what you are trying to set. If the open loop gain at some frequency is 10 and you want the closed loop gain to be 20 (set by your external resistors), the most you can get out is 10...
     
  10. Nov 8, 2017 #9
    when i do the calculation: Av = 1 + 2R/Rg which i have to "calculate overall circuit gain" is this "overall circuit gain" the closed loop gain?
     
  11. Nov 8, 2017 #10

    berkeman

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    Yes. When you analyze the opamp circuit that has negative feedback, you can initially make the assumption that the opamp Aol is infinite. That gives you the "virtual ground" property (where the - input is held at the same potential as the + input by the feedback) which helps you to solve for the Acl from the resistor values. You can then refine that by putting in the real Aol numbers to calculate the slight errors you get from the finite Aol.
     
  12. Nov 8, 2017 #11

    donpacino

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    Gold Member

    I realize this is a homework help forum, but just a quick comment. This is where, in real life, an additional amplifier stage would have to be added (or a change in amplifier) to get your desired closed loop gain.
     
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